# How do you use trigonometric substitution to write the algebraic expression sqrt(x^2+25) as a trigonometric function of theta where 0<theta<pi/2 and x=5tantheta?

Jun 24, 2017

$g \left(\theta\right) = 5 \sec \theta$

#### Explanation:

Let $\sqrt{{x}^{2} + 25}$

= $\sqrt{{\left(5 \tan \theta\right)}^{2} + 25}$

= $\sqrt{25 {\tan}^{2} \theta + 25}$

= $\sqrt{25 \left({\tan}^{2} \theta + 1\right)}$

= $5 \sqrt{{\sec}^{2} \theta}$

= $5 \sec \theta$

Hence $f \left(x\right) = \sqrt{{x}^{2} + 25}$ can be written as $g \left(\theta\right) = 5 \sec \theta$